The way to solve the missing side of a right triangle is normally by the Pythagorean Theorem:
[tex] {a}^{2} + {b}^{2} = {c}^{2} \\ {a}^{2} = {c}^{2} - {b}^{2} \: \infty \: \sqrt{ {a}^{2} } = \sqrt{( {c}^{2} - {b}^{2}) } \\ a= \sqrt{( {c}^{2} - {b}^{2}) } [/tex]
where c always is the hypotenuse
So c = 24 ft
b = 44 + a
[tex] {a}^{2} + {b}^{2} = {c}^{2} \\ {a}^{2} + {(a + 44)}^{2} = {24}^{2} \\ {a}^{2} + {a}^{2} + 88a + 1936 = 576[/tex]
Now combine like terms:
[tex] 2{a}^{2} + 88a + 1936 = 576 \\ 2{a}^{2} + 88a + 1936 - 576 = 0 \\ 2{a}^{2} + 88a + 1360 = 0[/tex]
Now we have to try to factor this quadratic equation, first let's take out 2:
[tex]2({a}^{2} + 44a + 680) = 0[/tex]
we need factors of 680 whose sum = 44
20×34, 10×68, 17×40
however... 20+34=54, 10+68=78, 17+40=57
So we will need to use the quadratic equation unfortunately :(
[tex]x = ( - b + - \sqrt{( {b}^{2} - 4ac)}) \div 2a [/tex]
Now the "x" is actually our "a", the a is 1, b is 44, c is 680
[tex]a = ( - 44 + - \sqrt{( {44}^{2} - 4(1)680)}) \\ \div \: 2(1) \\ a = ( - b + - \sqrt{(1936 - 2720)}) \\ \div \: 2 [/tex]
[tex]a = ( - 44 + - \sqrt{(-784)}) \\ \div \: 2 = - 44 \div 2 \: + - (i \sqrt{784}) \div 2 \\ a = - 22 + - i \sqrt{16} \sqrt{49} \div 2 \\ [/tex]
[tex]a = - 22 + - 28i \div 2 \\ a = - 22 + - 14i [/tex]
not quite sure where to go from these imaginary numbers...
I DON'T THINK A RIGHT TRIANGLE WITH THOSE DIMENSIONS IS POSSIBLE
